Finite-state Feynman-Kac evolution for continuous jump rates

If $T=0$, the single endpoint value is $G_y(0)={\rho }_y$. Suppose $T>0$ and put $K:={\max }_x{\sup }_{s\in [0,T]}\mathrm{\mid }{V}_x(s)\mathrm{\mid }$ and $M:={\max }_x{\sup }_{s\in [0,T]}{\lambda }_x(s)$. Write $w(t):=\exp (-{\int }_0^t{V}_{X_s}(s)ds)$. Almost surely the path has finitely many jumps by [F2], so the integrand is piecewise continuous; its integral is the limit of coordinate-measurable Riemann sums. Thus the weights are measurable (assign any bounded value on the measurable exceptional null set) and satisfy $0<w(t)\le e^{KT}$. For $0\le u<v\le T$ and $h:=v-u$, $\mathrm{\mid }w(v)-w(u)\mathrm{\mid }\le e^{KT}(e^{Kh}-1)$, while [F1] gives $\mathbb{P}(N_{(u,v]}\ge 1)\le 1-e^{-Mh}\le Mh$. Since an endpoint indicator can change only if a jump occurs, $\mathrm{\mid }{G}_y(v)-G_y(u)\mathrm{\mid }\le e^{KT}(e^{Kh}-1+Mh)\mathrm{.}$ This bound tends to zero with $h$, uniformly in $u,v$, proving continuity on the entire closed interval.

For $0\le t\le T-h$, condition on the history through $t$. Starting at $y$, the contribution of no jump to the additional weighted endpoint indicator is $\exp (-{\int }_t^{t+h}({\lambda }_y(s)+V_y(s))ds)=1-({\lambda }_y(t)+V_y(t))h+o(h)$. Starting at $x\mathrm{\ne }y$, exactly one jump to $y$ contributes ${\int }_t^{t+h}{H}_x(t,s){\kappa }_{xy}(s)H_y(s,t+h)\exp (-{\int }_t^s{V}_x(r)dr-{\int }_s^{t+h}{V}_y(r)dr)ds={\kappa }_{xy}(t)h+o(h)\mathrm{.}$ The conditional bound from [F2], obtained by restarting the first-jump rule at $t$, bounds the probability of two or more jumps by $(Mh{)}^2$ when $Mh<1$; their weighted contribution is at most $e^{Kh}(Mh{)}^2$. The finitely many continuous rates and potentials have a common modulus of continuity $\omega (h)\to 0$ on $[0,T]$. Hence the preceding remainders are $O(h\omega (h)+h^2)$ uniformly in the starting time and state. Multiply by $w(t)$ and average, using the Markov property and the bound in step 1.1, to obtain $G_y(t+h)=G_y(t)+h{R}_y(t)+r_y(t,h),R_y(t):={\sum }_{x\mathrm{\ne }y}{G}_x(t){\kappa }_{xy}(t)-G_y(t){\lambda }_y(t)-V_y(t)G_y(t),$ where ${\sup }_{0\le t\le T-h,y\in E}\mathrm{\mid }{r}_y(t,h)\mathrm{\mid }\mathrm{/}h\to 0$.

By step 1.1 and continuity of the rates and potentials, $R$ is continuous. Step 2.1 gives the right derivative $G_y^{\mathrm{\prime }}(t+)=R_y(t)$ for $t<T$. For $t>0$, apply its uniform expansion at $t-h$ to get $(G_y(t)-G_y(t-h))\mathrm{/}h=R_y(t-h)+r_y(t-h,h)\mathrm{/}h\to R_y(t)$. Thus the full derivative exists on $(0,T)$, extends continuously as the stated one-sided derivatives at the endpoints, and satisfies the claimed equation. The defining expectation at zero gives $G_y(0)={\rho }_y$.

If $G$ and $\tilde{G}$ are two solutions with the same initial vector, their difference $D$ satisfies $D(0)=0$ and $D(t)={\int }_0^{t}A(s)D(s)ds$ by [F5], applied componentwise, where $A(s)=Q(s{)}^{\mathsf{T}}-\mathrm{diag}(V(s))$ in column convention. Its coefficients and $D$ are continuous on $[0,T]$. Define $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}:={\max }_y\mathrm{\mid }{z}_y\mathrm{\mid }$, $B:={\sup }_s{\max }_y{\sum }_x\mathrm{\mid }{A}_{yx}(s)\mathrm{\mid }<\mathrm{\infty }$, and $C:={\sup }_s\mathrm{\parallel }D(s){\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. Then $\mathrm{\parallel }A(s)z{\mathrm{\parallel }}_{\mathrm{\infty }}\le B\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$. Starting from $\mathrm{\parallel }D(t){\mathrm{\parallel }}_{\mathrm{\infty }}\le C$, induction in the integral equation gives $\mathrm{\parallel }D(t){\mathrm{\parallel }}_{\mathrm{\infty }}\le C(Bt{)}^n\mathrm{/}n!$ for every $n\ge 1$; the induction integrates $s^n$, whose primitive is $s^{n+1}\mathrm{/}(n+1)$, by [F5]. For each fixed $t$ these bounds tend to zero, since successive terms have ratio $Bt\mathrm{/}(n+1)$, eventually at most $1\mathrm{/}2$. Hence $D=0$ and the solution is unique.